# Restriction of smooth functions.

Consider the following question:

Suppose that $X$ is a subset of $\mathbb{R}^n$ and $Z$ is a subset of $X$. Show that the restriction to $Z$ of any smooth map on $X$ is a smooth map on $Z$.

(Note: A smooth function is defined to be one that has continuous partial derivatives of all orders)

What exactly is this question asking? I don't see why this does not follow immediately, and so I must be missing something quite crucial. I don't think the question was intended to prompt an answer of the form

Proof: QED.

• How is "smooth map on $Y$" defined? Commented Sep 2, 2013 at 15:17
• @DanielFischer Has continuous partial derivatives of all orders. I will add this information to the question. Commented Sep 2, 2013 at 15:18
• Hmm, how are isolated points treated? Commented Sep 2, 2013 at 15:19
• @DanielFischer I'm not sure, they aren't mentioned. This is question 2 of chapter 1 from Differential Topology by Guillemin and Pollack. Perhaps there is some convention in this field? Commented Sep 2, 2013 at 15:22

For arbitrary sets $X \subset \mathbb{R}^{m}, Y \subset \mathbb{R}^{n}$, a function $f: X \to Y$ is, by definition, smooth, if for any $x \in X$ there exists an open neighborhood $x \in U \subset \mathbb{R}^m$ and a smooth function $F: U \to \mathbb{R}^n$ s.t. $F_{|U \cap X} = f_{|U \cap X}$.
So if $z \in Z \subset X$ and $f$ is smooth, then for any $z \in Z$ there exists an open neighborhood $z \in U \subset \mathbb{R}^m$ and a smooth function $F: U \to \mathbb{R}^n$ s.t. $F_{|U \cap X} = f_{|U \cap X}$. Hence, $f_{|Z}$ is smooth.
It really is straightforward. Write $f$ for your smooth map on $X$. You simply need to show that there is an open set $U$ containing $Z$, and a smooth extension of $f$ to $U$. (Incidentally this handles the issue of how to treat isolated points mentioned in the comments.)